Problem: Graph this system of equations and solve. $14x+2y = 6$ $-6x-2y = 2$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $14x+2y = 6$ , to slope-intercept form. $y = -7 x + 3$ The y-intercept for the first equation is $3$ , so the first line must pass through the point $(0, 3)$ The slope for the first equation is $-7$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move down (because it's negative) $1$ position to the right. $7$ positions down from $(0, 3)$ is $(1, -4)$ Graph the blue line so it passes through $(0, 3)$ and $(1, -4)$ Convert the second equation, $-6x-2y = 2$ , to slope-intercept form. $y = -3 x - 1$ The y-intercept for the second equation is $-1$ , so the second line must pass through the point $(0, -1)$ The slope for the second equation is $-3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move down (because it's negative) $1$ position to the right. $3$ positions down from $(0, -1)$ is $(1, -4)$ Graph the green line so it passes through $(0, -1)$ and $(1, -4)$ The solution is the point where the two lines intersect. The lines intersect at $(1, -4)$.